# How to find the limit $\lim\limits_{x\to1/\sqrt2}\frac{\cos(\sin^{-1}x)-x}{1-\cot(\cos^{-1}x)}$?

Find the limit $$\lim_{x\to \frac{1}{\sqrt2}}\frac{\cos(\sin^{-1}x)-x}{1-\cot(\cos^{-1}x)}$$

First, I tried solving this sum using a substitute $$\sin^{-1}x=t$$, So when $$x=\frac{1}{\sqrt{2}}$$, $$t=\frac{\pi}{4}$$

$$\lim_{t\to \frac{\pi}{4}}\frac{\cos(t)-\sin(t)}{1-?}$$

But then I have trouble finding a value for $$\cos^{-1}x$$ in terms of t. Is my path correct? How should I proceed? Any hint would be highly appreciated... Thanks :)

P.S: I prefer solutions without using L'Hopital rule.

• you can use $sin^{-1}(x) + cos^{-1}(x)= \pi/2.$ I hope this helps. Jan 28, 2022 at 11:57
• Use \ in front of sin and cos to get $\sin,\cos$. Jan 28, 2022 at 11:57

If $$\sin^{-1}x=t\implies\cos^{-1}x=\frac\pi2-t$$, therefore, $$\lim_{t\to\frac\pi4}\frac{\cos t-\sin t}{1-\cot(\frac\pi2-t)}=\lim_{t\to\frac\pi4}\frac{\cos t-\sin t}{1-\tan t}=\cos\frac\pi4$$ Note that $$\theta=\sin^{-1}x$$ and $$\alpha=\cos^{-1}x$$.
According to image, can you find the value of $$\cos(\sin^{-1}x)$$ and $$\cot(\cos^{-1}x)$$?
We can find from the right triangle that: $$\cos(\arcsin(x))=\sqrt{1-x^{2}},\tag{1}$$ and $$\cot(\arccos(x))=\frac{x}{\sqrt{1-x^{2}}},\tag{2}$$ Therefore, $$\lim_{x\to \frac{1}{\sqrt{2}}}\frac{\cos(\arcsin(x))-x}{1-\cot(\arccos(x))}=\lim_{x\to \frac{1}{\sqrt{2}}}\frac{\sqrt{1-x^{2}}-x}{1-\frac{x}{\sqrt{1-x^{2}}}}=\lim_{x\to \frac{1}{\sqrt{2}}}(\sqrt{1-x^{2}})=\frac{1}{\sqrt{2}}.$$